Beats — Question 8

Question 8

A short averaging window retains the beat cross term

Two equal, nearly equal-frequency copropagating tones each have intensity $I_1$. After averaging over rapid carrier cycles, their combined intensity is $I(t)=2I_1[1+\cos(2\pi\Delta f\,t)]$, with phase alignment at $t=0$. Average this expression over $0\le t\le T$.

1. Derive the finite-window average.

2. Evaluate it for one full beat period and for one quarter of a beat period. Explain the long-window limit.

Question 8 - Solution

A short averaging window retains the beat cross term

(a) Method and calculation

Integrate the cross term explicitly:$$\overline I_T=\frac1T\int_0^T I(t)\,dt=2I_1\left[1+\frac{\sin(2\pi\Delta f T)}{2\pi\Delta f T}\right].$$The starting phase matters for short windows; the supplied expression specifies that phase.

(b) Interpretation and further calculation

For $T=1/\Delta f$, the sine term is zero and $\overline I_T=2I_1$. For $T=1/(4\Delta f)$,$$\overline I_T=2I_1(1+2/\pi)=3.27324I_1.$$As $T\Delta f\to\infty$, the bounded sine numerator divided by $T$ vanishes. Long averaging removes the cross term and recovers the sum of individual intensities.

Physical check

The carrier averaging assumes many fast cycles within the time window while still allowing the beat envelope to vary slowly.